8.09(F14) Chapter 1: A Review of Analytical Mechanics · PDF fileA Review of Analytical...

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Chapter 1 A Review of Analytical Mechanics 1.1 Introduction These lecture notes cover the third course in Classical Mechanics, taught at MIT since the Fall of 2012 by Professor Stewart to advanced undergraduates (course 8.09) as well as to graduate students (course 8.309). In the prerequisite classical mechanics II course the students are taught both Lagrangian and Hamiltonian dynamics, including Kepler bound motion and central force scattering, and the basic ideas of canonical transformations. This course briefly reviews the needed concepts, but assumes some familiarity with these ideas. References used for this course include Goldstein, Poole & Safko, Classical Mechanics, 3rd edition. Landau and Lifshitz vol.6, Fluid Mechanics. Symon, Mechanics for reading material on non-viscous fluids. Strogatz, Nonlinear Dynamics and Chaos. Review: Landau & Lifshitz vol.1, Mechanics. (Typically used for the prerequisite Classical Mechanics II course and hence useful here for review) 1.2 Lagrangian & Hamiltonian Mechanics Newtonian Mechanics In Newtonian mechanics, the dynamics of a system of N particles are determined by solving for their coordinate trajectories as a function of time. This can be done through the usual vector spatial coordinates r i (t) for i ∈{1,...,N }, or with generalized coordinates q i (t) for i ∈{1,..., 3N } in 3-dimensional space; generalized coordinates could be angles, et cetera. 1

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Chapter 1

A Review of Analytical Mechanics

1.1 Introduction

These lecture notes cover the third course in Classical Mechanics, taught at MIT sincethe Fall of 2012 by Professor Stewart to advanced undergraduates (course 8.09) as well asto graduate students (course 8.309). In the prerequisite classical mechanics II course thestudents are taught both Lagrangian and Hamiltonian dynamics, including Kepler boundmotion and central force scattering, and the basic ideas of canonical transformations. Thiscourse briefly reviews the needed concepts, but assumes some familiarity with these ideas.References used for this course include

• Goldstein, Poole & Safko, Classical Mechanics, 3rd edition.

• Landau and Lifshitz vol.6, Fluid Mechanics. Symon, Mechanics for reading materialon non-viscous fluids.

• Strogatz, Nonlinear Dynamics and Chaos.

• Review: Landau & Lifshitz vol.1, Mechanics. (Typically used for the prerequisiteClassical Mechanics II course and hence useful here for review)

1.2 Lagrangian & Hamiltonian Mechanics

Newtonian Mechanics

In Newtonian mechanics, the dynamics of a system of N particles are determined by solvingfor their coordinate trajectories as a function of time. This can be done through the usualvector spatial coordinates ri(t) for i ∈ {1, . . . , N}, or with generalized coordinates qi(t) fori ∈ {1, . . . , 3N} in 3-dimensional space; generalized coordinates could be angles, et cetera.

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Velocities are represented through vi ≡ ri for spatial coordinates, or through qi forgeneralized coordinates. Note that dots above a symbol will always denote the total timederivative d . Momenta are likewise either Newtonian pi = mivi or generalized pi.dt

For a fixed set of masses m Newton’s 2ndi law can be expressed in 2 equivalent ways:

1. It can be expressed as N second-order equations Fi = d (miri) with 2N boundarydt

conditions given in ri(0) and ri(0). The problem then becomes one of determining theN vector variables ri(t).

2. It can also be expressed as an equivalent set of 2N 1st order equations Fi = pi &pi/mi = ri with 2N boundary conditions given in ri(0) and pi(0). The problem thenbecomes one of determining the 2N vector variables ri(t) and pi(t).

Note that F = ma holds in inertial frames. These are frames where the motion of aparticle not subject to forces is in a straight line with constant velocity. The converse does nothold. Inertial frames describe time and space homogeneously (invariant to displacements),isotropically (invariant to rotations), and in a time independent manner. Noninertial framesalso generically have fictitious “forces”, such as the centrifugal and Coriolis effects. (Inertialframes also play a key role in special relativity. In general relativity the concept of inertialframes is replaced by that of geodesic motion.)

The existence of an inertial frame is a useful approximation for working out the dynam-ics of particles, and non-inertial terms can often be included as perturbative corrections.Examples of approximate inertial frames are that of a fixed Earth, or better yet, of fixedstars. We can still test for how noninertial we are by looking for fictitious forces that (a) maypoint back to an origin with no source for the force or (b) behave in a non-standard fashionin different frames (i.e. they transform in a strange manner when going between differentframes).

We will use primes will denote coordinate transformations. If r is measured in an inertialframe S, and r′ is measured in frame S ′ with relation to S by a transformation r′ = f(r, t),then S ′ is inertial iff r = 0 ↔ r′ = 0. This is solved by the Galilean transformations,

r′ = r + v0t

t′ = t,

which preserves the inertiality of frames, with F = mr and F′ = mr′ implying each other.Galilean transformations are the non-relativistic limit, v � c, of Lorentz transformationswhich preserve inertial frames in special relativity. A few examples related to the conceptsof inertial frames are:

1. In a rotating frame, the transformation[ is given by

x′

y′

]=

[cos(θ) sin(θ) x− sin(θ) cos(θ)

] [y

]If θ = ωt for some constant ω, then r = 0 still gives r′ 6= 0, so the primed frame isnoninertial.

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Figure 1.1: Frame rotated by an angle θ

2. In polar coordinates, r = rr, gives

dr ˆdθ˙ˆ= θθ,dt

˙= −θr (1.1)dt

and thus¨ ˙ˆ ¨ˆ ˙r = rr + 2rθθ + r θθ − θ2r . (1.2)

Even if ¨ ¨r = 0 we can still have r 6= 0 and θ

( )6= 0, and we can not in general form

a simple Newtonian force law equation mq = Fq for each of these coordinates. Thisis different than the first example, since here we are picking coordinates rather thanchanging the reference frame, so to remind ourselves about their behavior we will callthese ”non-inertial coordinates” (which we may for example decide to use in an inertialframe). In general, curvilinear coordinates are non-inertial.

Lagrangian Mechanics

In Lagrangian mechanics, the key function is the Lagrangian

L = L(q, q, t). (1.3)

Here, q = (q1, . . . , qN) and likewise q = (q1, . . . , qN). We are now letting N denote thenumber of scalar (rather than vector) variables, and will often use the short form to denotedependence on these variables, as in Eq. (1.3). Typically we can write L = T − V whereT is the kinetic energy and V is the potential energy. In the simplest cases, T = T (q)and V = V (q), but we also allow the more general possibility that T = T (q, q, t) andV = V (q, q, t). It turns out, as we will discuss later, that even this generalization does notdescribe all possible classical mechanics problems.

The solution to a given mechanical problem is obtained by solving a set of N second-orderdifferential equations known as Euler-Lagrange equations of motion,

d ∂

dt

(L ∂

∂qi

)L− = 0. (1.4)∂qi

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

These equations involve qi, and reproduce the Newtonian equations F = ma. The principleof stationary action (Hamilton’s principle),

t2

δS = δ

∫L(q, q, t) dt = 0, (1.5)

t1

is the starting point for deriving the Euler-Lagrange equations. Although you have coveredthe Calculus of Variations in an earlier course on Classical Mechanics, we will review themain ideas in Section 1.5.

There are several advantages to working with the Lagrangian formulation, including

1. It is easier to work with the scalars T and V rather than vectors like F.

2. The same formula in equation (1.4) holds true regardless of the choice of coordinates.To demonstrate this, let us consider new coordinates

Qi = Qi(q1, . . . , qN , t). (1.6)

This particular sort of transformation is called a point transformation. Defining thenew Lagrangian by

L′ = L′ ˙(Q,Q, t) = L(q, q, t), (1.7)

we claim that the equations of motion are simply

d ∂

dt

(L′ ∂˙∂Qi

)L′− = 0. (1.8)

∂Qi

Proof: (for N = 1, since the generalization is straightforward)Given L′ ˙(Q,Q, t) = L(q, q, t) with Q = Q(q, t) then

dQ =

∂QQ(q, t) =

dt

∂Qq +

∂q. (1.9)

∂t

Therefore

˙∂Q ∂Q=

∂q, (1.10)

∂q

a result that we will use again in the future. Then

∂L ∂L′=

∂q

∂L′=

∂q

∂Q

∂Q

∂L′+

∂q

˙∂Q˙∂Q

, (1.11)∂q

∂L ∂L′=

∂q

∂L′=

∂q

˙∂Q˙∂Q

∂L′=

∂q

∂Q˙∂Q

.∂q

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Since ∂Q ′= 0 there is no term ∂L

∂q∂Q

∂Qin the last line.

∂q

Plugging these results into 0 = d ∂dt

(L ∂∂q

)− L gives

∂q

0 =

[d(∂L′

dt

∂˙∂Q

)Q ∂L′

+∂q

d˙∂Q

dt

(Q ∂

∂q

)]−[L′ ∂Q

∂Q

∂L′+

∂q

˙∂Q˙∂Q[ ∂q

d

]=

dt

(L′ ∂˙∂Q

)L′− ∂

∂Q

]Q, (1.12)

∂q

since d ∂Qdt

= (q ∂∂q

+ ∂∂q

)∂Q∂t

= ∂∂q

(q ∂∂q

+ ∂∂q

˙)Q = ∂Q

∂tso that the second and fourth terms

∂q

cancel. Finally for non-trivial transformation where ∂Q =∂q6 0 we have, as expected,

d0 =

dt

(L′ ∂˙∂Q

)L′− . (1.13)∂Q

Note two things:

• This implies we can freely use the Euler-Lagrange equations for noninertial coor-dinates.

• We can formulate L in whatever coordinates are easiest, and then change toconvenient variables that better describe the symmetry of a system (for example,Cartesian to spherical).

3. Continuing our list of advantages for using L, we note that it is also easy to incorporateconstraints. Examples include a mass constrained to a surface or a disk rolling withoutslipping. Often when using L we can avoid discussing forces of constraint (for example,the force normal to the surface).

Lets discuss the last point in more detail (we will also continue to discuss it in the nextsection). The method for many problems with constraints is to simply make a good choice forthe generalized coordinates to use for the Lagrangian, picking N − k independent variablesqi for a system with k constraints.

Example: For a bead on a helix as in Fig. 1.2 we only need one variable, q1 = z.

Example: A mass m2 attached by a massless pendulum to a horizontally sliding mass m1

as in Fig. 1.3, can be described with two variables q1 = x and q2 = θ.

Example: As an example using non-inertial coordinates consider a potential V = V (r, θ)˙ˆin polar coordinates for a fixed mass m at position r = rr. Since r = rr + rθθ we have

T = m r2 = m2

˙r2

(2 + r2θ2

), giving

mL = ˙r

2

(2 + r2θ2

)− V (r, θ). (1.14)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Figure 1.2: Bead on a helix

Figure 1.3: Pendulum of mass m2 hanging on a rigid bar of length ` whose support m1 is africtionless horizontally sliding bead

For r the Euler-Lagrange equation is

d0 =

dt

(L ∂

∂r

)L− d

=∂r

∂˙(mr)dt

−mrθ2 V+ . (1.15)∂r

This gives

mr − ˙mrθ2 ∂V= − = Fr, (1.16)

∂r

from which we see that Fr 6= mr. For θ the Euler-Lagrange equation is

d0 =

dt

(L ∂˙∂θ

)L− d

=∂θ

∂˙mrdt

(2θ) V

+ . (1.17)∂θ

This givesd ∂˙mrdt

(2θ) V

= − = Fθ, (1.18)∂θ

which is equivalent to the relation between angular momentum and torque perpendicular to˙the plane, Lz = Fθ = τz. (Recall L = r× p and τ = r× F.)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Figure 1.4: Particle on the inside of a cone

Example: Let us consider a particle rolling due to gravity in a frictionless cone, shownin√ Fig. 1.4, whose opening angle α defines an equation for points on the cone tan(α) =x2 + y2/z. There are 4 steps which we can take to solve this problem (which are more

general than this example):

1. Formulate T and V by N = 3 generalized coordinates. Here it is most convenient to

choose cylindrical coordinates denoted (r, θ, z), so that T = m ˙r2 + r2θ2 + z2 and2

V = mgz.

( )2. Reduce the problem to N − k = 2 independent coordinates and determine the new

Lagrangian L = T − V . In this case we eliminate z = r cot(α) and z = r cot(α), so

mL = ˙1

2

[(+ cot2 α

)r2 + r2θ2

]−mgr cotα. (1.19)

3. Find the Euler-Lagrange equations. For r, 0 = d ∂dt

(L ∂∂r

)− L , which here is

∂r

d0 = ˙

dt

[m(1 + cot2 α

)r]−mrθ2 +mg cotα (1.20)

giving (1 + cot2 ˙α

)r − rθ2 + g cotα = 0. (1.21)

For θ we have 0 = d ∂dt

(L ∂˙∂θ

)− L , so

∂θ

d0 = ,

dt

(mr2θ

)− 0 (1.22)

giving˙ ¨(2rθ + rθ)r = 0. (1.23)

4. Solve the system analytically or numerically, for example using Mathematica. Or wemight be only interested in determining certain properties or characteristics of themotion without a full solution.

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Hamiltonian Mechanics

In Hamiltonian mechanics, the canonical momenta pi ≡ ∂L are promoted to coordinates∂qi

on equal footing with the generalized coordinates qi. The coordinates (q, p) are canonicalvariables, and the space of canonical variables is known as phase space.

The Euler-Lagrange equations say pi = ∂L . These need not equal the kinematic momenta∂qi

miqi if V = V (q, q). Performing the Legendre transformation

H(q, p, t) = qipi − L(q, q, t) (1.24)

(where for this equation, and henceforth, repeated indices will imply a sum unless otherwisespecified) yields the Hamilton equations of motion

∂Hqi = (1.25)

∂pi∂H

pi = −∂qi

which are 2N 1st order equations. We also have the result that

∂H ∂L=

∂t− . (1.26)∂t

Proof: (for N = 1) Consider

∂HdH =

∂Hdq +

∂q

∂Hdp+

∂pdt (1.27)

∂t

=∂L

pdq + qdp− ∂L��dq

∂q− �

∂Ldq

∂q− dt . (1.28)∂t

Since we are free to independently vary dq, dp, and dt this implies ∂L = p, ∂L∂q

= p, and∂q

∂H =∂t

−∂L .∂t

We can interpret the two Hamilton equations as follows:

• qi = ∂H is an inversion of pi = ∂L∂pi

= pi(q, q, t).∂qi

• pi = −∂H provides the Newtonian dynamics.∂qi

However, these two equation have an have equal footing in Hamiltonian mechanics, sincethe coordinates and momenta are treated on a common ground. We can use pi = ∂L to

∂qiconstruct H from L and then forget about L.

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

As an example of the manner in which we will usually consider transformations betweenLagrangians and Hamiltonians, consider again the variables relevant for the particle on acone from Fig. 1.4:

˙ z=r cotα ˙L(r, θ, z, r, θ, z) −→ new L(r, θ, r, θ) −→ Euler-Lagrange Eqtns. (1.29)

l l lnot here

H(r, θ, z, pr, pθ, pz) =⇒ H(r, θ, pr, pθ) −→ Hamilton Eqtns.

Here we consider transforming between L and H either before or after removing the redun-dant coordinate z, but in this course we will only consider constraints imposed on Lagrangiansand not in the Hamiltonian formalism (the step indicated by =⇒). For the curious, the topicof imposing constraints on Hamiltonians, including even more general constraints than thosewe will consider, is covered well in Dirac’s little book “Lectures on Quantum Mechanics”.Although Hamiltonian and Lagrangian mechanics provide equivalent formalisms, there is of-ten an advantage to using one or the other. In the case of Hamiltonian mechanics potentialadvantages include the language of phase space with Liouville’s Theorem, Poisson Bracketsand the connection to quantum mechanics, as well as the Hamilton-Jacobi transformationtheory (all to be covered later on).

Special case: Let us consider a special case that is sufficient to imply that the Hamiltonianis equal to the energy, H = E ≡ T + V . If we only have quadratic dependence on velocitiesin the kinetic energy, T = 1Tjk(q)qj qk, and V = V (q) with L = T V , then

2−

∂Lqipi = qi

1=

∂qi

1qiTikqk +

2qjTjiqi = 2T. (1.30)

2

Hence,H = qipi − L = T + V = E (1.31)

which is just the energy.

Another Special case: Consider a class of Lagrangians given as

1L(q, q, t) = L0 + aj qj + qjTjkqk (1.32)

2

where L0 = L0(q, t), aj = aj(q, t), and Tjk = Tkj = Tjk(q, t). We can write this in shorthandas

1L = L0 ~a · ˙+ ~q + ~q

2·T · ~q . (1.33)

Here the generalized coordinates, momenta, and coefficients have been collapsed into vectors,like ~q (rather than the boldface that we reserve for Cartesian vectors), and dot products of

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

vectors from the left imply transposition of that vector. Note that ~q is an unusual vector,since its components can have different dimensions, eg. ~q = (x, θ), but nevertheless thisnotation is useful. To find H,

∂Lpj = = aj + Tjkqk, (1.34)

∂qjˆmeaning p~ = ~a + T · ˙ ˙ ˆ~q. Inverting this gives ~q = T−1· (p~ − ˆ~a), where T−1 will exist because

ˆof the positive-definite nature of kinetic energy, which implies that T is a postive definite˙matrix. Thus, H = ~q · p~− L yields

1H = − ˆ(p~ ~a) · T−1 · (p~− ~a)

2− L0(q, t) (1.35)

ˆas the Hamiltonian. So for any Lagrangian in the form of Eq. (1.32), we can find T−1 andwrite down the Hamiltonian as in Eq. (1.35) immediately.

Example: let us consider L = 1mv2− eφ+ eA · v, where e is the electric charge and SI2

ˆunits are used. In Eq. (1.32), because the coordinates are Cartesian, a = eA, T = m1, andL0 = −eφ, so

1H = (p +

m− eA)2 eφ . (1.36)

2As you have presumably seen in an earlier course, this Hamiltonian does indeed reproducethe Lorentz force equation e(E + v ×B) = mv.

A more detailed Example. Find L and H for the frictionless pendulum shown in Fig. 1.3.This system has two constraints, that m1 is restricted to lie on the x-axis sliding withoutfriction, and that the rod between m1 and m2 is rigid, giving

y1 = 0 , (y1 − y2)2 + (x1 − x2)2 = `2 . (1.37)

Prior to imposing any constraints the Lagrangian ism

L = T − 1V =

mx2 2

2 1 + (x22 + y2

2)−m2gy22

−m1gy1 . (1.38)

Lets choose to use x ≡ x1 and the angle θ as the independent coordinates after imposingthe constraints in Eq. (1.37). This allows us to eliminate y1 = 0, x2 = x + ` sin θ andy2 = − ˙ ˙` cos θ, together with x2 = x + ` cos θ θ, y2 = ` sin θ θ, x1 = x. The Lagrangian withconstraints imposed is

m1L =

mx2 2

+2

(x2 ˙ ˙+ 2` cos θ xθ + `2 cos2 ˙θ θ2 + `2 sin2 θ θ2

2

)+m2g` cos θ . (1.39)

Next we determine the Hamiltonian. First we find

∂Lpx = ˙ ˙= m1x+m2(x+ ` cos θθ) = (m1 +m2)x+m2` cos θ θ , (1.40)

∂x∂L

pθ = ˙= m˙ 2` cos θ x+m2`

2θ .∂θ

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Note that px is(not)simply pr(oportional to x here (actually px is the center-of-mass momen-px xˆtum). Writing = Tpθ

·θ

)gives

T =

(m1 +m2 m2` cos θ

,m2` cos θ m2`

2

)(1.41)

1with L =

x˙ ˆ(x θ) T + L0 where L0 = m˙ 2g` cos θ. Computing2

· ·(θ

)

T−1 1=

m

m1m2`2 +m2`2 sin2 θ

(2`

2 −m2` cos θ, (1.42)−m2` cos θ m1 +m2

)we can simply apply Eq. (1.35) to find the corresponding Hamiltonian

1H =

pˆ(px pθ)2

·T−1·(

x

)−m2g` cos θ (1.43)

1= m `2

2 p2 + (m1 +m2)p2 2m2` cos θpxpθ m2g` cos θ .2m2`2(m 2 x θ

1 +m2 sin θ)

[−

]−

Lets compute the Hamilton equations of motion for this system. First for (x, px) we find

∂Hx =

px=

∂px

cos θ p

m1 +m2 sin2 θ− θ

,`(m1 +m2 sin2 (1.44)

θ)

∂Hpx = − = 0 .

∂x

As we might expect, the CM momentum is time independent. Next for (θ, pθ):

∂Hθ =

1=

∂pθ(m1 +m2)pθ m2` cos θpx , (1.45)

m2`2(m1 +m2 sin2 θ)

[−

∂H

]pθ = − sin θ cos θ

=∂θ

m2`2p2 + (m m2)p2

1 + 2m2` cos θpxpθ` 1 +m sin2 θ)2

[x2(m θ

2

sin θp− θ

]m2g` sin − xp

θ .`(m1 +m2 sin θ)

These non-linear coupled equations are quite complicated, but could be solved in math-ematica or another numerical package. To test our results for these equations of motionanalytically, we can take the small angle limit, approximating sin θ ' θ, cos θ ' 1 to obtain

pxx =

p

m1

− θ 1˙, px = 0 , θ =`m1

(m1m2`2

[m1 +m2)pθ −m2`px

],

θpθ =

[ θpxpθm2`

2p2x + (m1 +m2)p2

θ − 2m2` cos θpxpθ`2m2

1

]− m

`m1

− 2g`θ . (1.46)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

To simplify it further we can work in the CM frame, thus setting px = 0, and linearize theequations by noting that pθ ∼ θ should be small for θ to remain small, and hence θp2

θ is ahigher order term. For the non-trivial equations this leaves

px = − θ p˙ θ

, θ =`m1

, pθ =µ`2

−m2g`θ , (1.47)

¨where µ = m1m2/(m1 + m2) is the reduced mass for the two-body system. Thus θ =pθ/(µ`

2) = −m2 g θ as expected for simple harmonic motion.µ `

1.3 Symmetry and Conservation Laws

A cyclic coordinate is one which does not appear in the Lagrangian, or equivalently in theHamiltonian. Because H(q, p, t) = qipi − L(q, q, t), if qj is absent in L for some particular j,it will be absent in H as well. The absence of that qj corresponds with a symmetry in thedynamics.

In this context, Noether’s theorem means that a symmetry implies a cyclic coordinate,which in turn produces a conservation law. If qj is a cyclic coordinate for some j, thenwe can change that coordinate without changing the dynamics given by the Lagrangianor Hamiltonian, and hence there is a symmetry. Furthermore the corresponding canonicalmomentum pj is conserved, meaning it is a constant through time.

The proof is simple. If ∂L = 0 then pj = d∂qj

∂Ldt

= ∂L∂qj

= 0, or even more simply, ∂H∂qj

= 0∂qj

is equivalent to pj = 0, so pj is a constant in time.Special cases and examples of this abound. Lets consider a few important ones:

1. Consider a system of N particles where no external or internal force acts on the centerof mass (CM) coordinate R = 1 miri, where the total mass M =

M imi. Then theCM momentum P is conserved. This is because

∑FR = −∇RV = 0 (1.48)

so V is independent of R. Meanwhile, T = 1 m 2i ir2 i , which when using coordinates

relative to the center of mass, r′i ≡ ri −R, bec

∑omes

1T =

(d˙m

2

∑i

i

)R2 ˙+R ·

(1

mdt

∑ir′i

i

)+

∑ 1mir

′22 i =

i

1˙MR2 +2

(1.49)2

∑mir

′2i .

i

Note that∑

imir′i = 0 from the definitions of M , R, and r′i, so T splits into two

terms, one for the CM motion and one for relative motion. We also observe that T is˙independent of R. This means that R is cyclic for the full Lagrangian L, so P = MR

is a conserved quantity. In our study of rigid bodies we will also need the forms of Mand R for a con∫tinuous body with mass distribution ρ(r), which for a three dimensionalbody are M = d3r ρ(r) and R = 1 d

M

∫3r ρ(r) r.

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Note that P = 0 is satisfied by having no total external force, so Fext = i Fexti =

0, and by the internal forces obeying Newton’s 3rd law Fi→j = −Fj→i. Hence,

∑¨MR =

∑Fexti +

∑Fi j = 0. (1.50)→

i i,j

2. Let us consider a system that is invariant with respect to rotations of angle φ about asymmetry axis. This has a conserved angular momentum. If we pick φ as a generalizedcoordinate, then L = T−V is independent of φ, so pφ = ∂L = 0 meaning pφ is constant.

∂φ

˙In particular, for a system where V is independent of the angular velocity φ we have

∂Tpφ =

∑ ∂r= miri

∂ϕi

· i ∂=

∂ϕ

∑ rmivi

i

· i. (1.51)

∂ϕ

Simplifying further using the results in Fig. 2.2 yields

pϕ =∑

mivi · (ni

× ri) = n ·∑

rii

×mivi = n · Ltotal. (1.52)

Figure 1.5: Rotation about a symmetry axis

Note∑ that L about the CM is conserved for systems with no external torque,τ ext = i ri×Fext

i = 0 and internal forces that are all central. Defining rij ≡ ri−rj andits magnitude appropriately, this means Vij = Vij(rij). This implies that Fji = −∇iVij(no sum on the repeated index) is parallel to rij. Hence,

dL=

dt

∑ri

i

× pi =∑

rii

× Fexti +

∑ri

i,j

× Fji. (1.53)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

However,∑

i ri × Fexti = 0, so

dL=∑

rij =i<j

× Fji 0. (1.54)dt

3. One can also consider a scaling transformation. Suppose that under the transformationri → λri the potential is homogeneous and transforms as V → λkV for some constantk. Letting T be quadratic in ri and taking time to transform as t λ1−k/2 t thengives r → λk/2 k

→i ri. So by construction T → λ T also, and thus the full Lagrangian

L→ λkL. This overall factor does not change the Euler-Lagrange equations, and hencethe transformation is a symmetry of the dynamics, only changing the overall scale orunits of the coordinate and time variables, but not their dynamical relationship. Thiscan be applied for several well known potentials:

a) k = 2 for a harmonic oscillator. Here the scaling for time is given by 1− k/2 = 0,so it does not change with λ. Thus, the frequency of the oscillator, which is atime variable, is independent of the amplitude.

b) k = −1 for the Coulomb potential. Here 1−k/2 = 3/2 so there is a more intricaterelation between coordinates and time. This power is consistent with the behaviorof bound state orbits, where the period of the orbit T obeys T 2 ∝ a3, for a thesemi-major axis distance (Kepler’s 3rd law).

c) k = 1 for a uniform gravitational field.√ Here 1− k/2 = 1/2 so for a freely fallingobject, the time of free fall goes as h where h is the distance fallen.

4. Consider the Lagrangian for a charge in electromagnetic fields, L = 1mr2−eφ+eA · r.2

As a concrete example, let us take φ and A to be independent of the Cartesian coor-dinate x. The canonical momentum is p = ∂L = mr + eA, which is notably different

∂r

from the kinetic momentum. Then x being cyclic means the canonical momentum pxis conserved.

5. Let us consider the conservation of energy and the relationship between energy and˙the Hamiltonian. Applying the time derivative gives H = ∂H q + ∂H

∂qp+ ∂H

∂p. However,

∂t

q = ∂H and p =∂p

−∂H . Thus∂q

∂HH =

∂L=

∂t− . (1.55)∂t

There are two things to consider.

• If H (or L) has no explicit time dependence, then H = qipi − L is conserved.

• ˙Energy is conserved if E = 0, where energy is defined by E = T + V .

If H = E then the two points are equivalent, but otherwise either of the two could betrue while the other is false.

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Example: Let us consider a system which provides an example where H = E butenergy is not conserved, and where H 6= E but H is conserved. The two situationswill be obtained from the same example by exploiting a coordinate choice. Consider asystem consisting of a mass m attached by a spring of constant k to a cart moving at aconstant speed v0 in one dimension, as shown in Fig. 1.6. Let us call x the displacement

Figure 1.6: Mass attached by a spring to a moving cart

of m from the fixed wall and x′ is its displacement from the center of the moving cart.Using x,

mL(x, x) = T − V =

kx2

2− (x− v0t)

2 , (1.56)2

where the kinetic term is quadratic in x and the potential term is independent of x.This means that H falls in the special case considered in Eq. (1.31) so

p2

H = E = T + V =k

+2m

(x2− v0t)

2 , (1.57)

However ∂H the∂6= 0 so the energy is not conserved. (Of course full energy would

t

be conserved, but we have not accounted for the energy needed to pull the cart at aconstant velocity, treating that instead as external to our system. That is what led tothe time dependent H.)

If we instead choose to use the coordinate x′ = x− v0t, then

mL′(x′, x′) =

mx′2 +mv0x

′ +2

kv2

2 0 − x′2. (1.58)2

Note that p′ = mx′+mv0 = mx = p. This Lagrangian fits the general form in equation(1.32) with a = mv 2 2

0 and L0 = mv0/2− kx′ /2. So

1H ′(x′, p′) = x′p′ − L′ = 2 k

(p′2m

−mv0) +m

x′22− v2

2 0, (1.59)

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Here the last terms is a constant shift. The first and second terms in this expression forH ′ look kind of like the energy that we would calculate if we were sitting on the cartand did not know it was moving, which is not the same as the energy above. Hence,H ′ 6 ˙ ′

= E, but H ′ = 0 because ∂H = 0, so H ed.t

′ is conserv∂

1.4 Constraints and Friction Forces

So far, we’ve considered constraints to a surface or curve that are relationships betweencoordinates. These fall in the category of holonomic constraints. Such constraints take theform

f(q1, . . . , qN , t) = 0 (1.60)

where explicit time dependence is allowed as a possibility. An example of holonomic constrainis mass in a cone (Figure 1.4), where the constrain is z−r cotα = 0. Constraints that violatethe form in Eq. (1.60) are non-holonomic constraints.

• An example of a non-holonomic constraint is a mass on the surface of a sphere. The

Figure 1.7: Mass on a sphere

constraint here is an inequality r2 − a2 ≥ 0 where r is the radial coordinate and a isthe radius of the sphere.

• Another example of a non-holonomic constraint is an object rolling on a surface with-out slipping. The point of contact is stationary, so the constraint is actually on thevelocities.

A simple example is a disk of radius a rolling down an inclined plane without slipping,˙as shown in Fig. 1.8. Here the condition on velocities, aθ = x is simple enough that it can

be integrated into a holonomic constraint.As a more sophisticated example, consider a vertical disk of radius a rolling on a horizontal

plane, as shown in Fig. 1.9. The coordinates are (x, y, θ, φ), where (x, y) is the point ofcontact, φ is the rotation angle about its own axis, and θ is the angle of orientation along thexy-plane. We will assume that the flat edge of the disk always remain parallel to z, so the

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Figure 1.8: Disk rolling down an incline without slipping

Figure 1.9: Vertical rolling disk on a two dimensional plane

˙disk never tips over. The no-slip condition is v = aφ where v is the velocity of the center ofthe disk, and v = | ˙v|. This means x = v sin(θ) = a sin(θ)φ and y = − ˙v cos(θ) = −a cos(θ)φ,or in differential notation, dx− a sin(θ)dφ = 0 and dy + a cos(θ)dφ = 0.

In general, constraints of the form∑aj(q)dqj + at(q)dt = 0 (1.61)

j

are not holonomic. We will call this a semi-holonomic constraint, following the terminologyof Goldstein.

Let us consider the special case of a holonomic constraint in differential form, f(q1, ..., q3N , t) =0. This means

ff =

∑ ∂d

j

∂fdqj +

∂qjdt = 0, (1.62)

∂t

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so aj = ∂f and at = ∂f∂qj

. The symmetry of mixed partial derivatives means∂t

∂aj ∂ai=

∂qi

∂at,

∂qj

∂ai=

∂qi. (1.63)

∂t

These conditions imply that a seemingly semi-holonomic constraint is in fact holonomic. (Inmath we would say that we have an exact differential form df for the holonomic case, butthe differential form in Eq.(1.61) need not always be exact.)

Example: To demonstrate that not all semiholonomic constrants are secretly holo-nomic, consider the constraint in the example of the vertical disk. Here there is no func-tion h(x, y, θ, φ) that we can multiply the constraint df = 0 by to make it holonomic.For the vertical disk from before, we could try (dx − a sin(θ) dφ)h = 0 with ax = h,

aφ = − ∂aa sin(θ)h, aθ = 0, and ay = 0 all for some function h. As we must have φ = ∂aθ

∂θ,

∂φ

then 0 = −a cos(θ)− a sin(θ)∂h , so h = k∂θ

. That said, ∂axsin(θ)

= ∂aθ∂θ

gives ∂h∂x

= 0 which is a∂θ

contradiction for a non-trivial h with k 6= 0.If the rolling is instead constrained to a line rather than a plane, then the constraint is

holonomic. Take as an example θ = π ˙for rolling along x, then x = aφ and y = 0. Integrating2

we have x = aϕ+x , y = y π0 0, and θ = , which together form a set of holonomic constraints.

2

A useful concept for discussing constraints is that of the virtual displacement δri ofparticle i. There are a few properties to be noted of δri.

• It is infinitesimal.

• It is consistent with the constraints.

• It is carried out at a fixed time (so time dependent constraints do not change its form).

Example: let us consider a bead constrained to a moving wire. The wire is oriented along

Figure 1.10: Bead on a moving wire

the x-axis and is moving with coordinate y = v0t. Here the virtual displacement of the

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bead δr is always parallel to x (since it is determined at a fixed time), whereas the realdisplacement dr has a component along y in a time interval dt.

For a large number of constraints, the constraint force Zi is perpendicular to δri, meaningZi · δri = 0, so the “virtual work” (in analogy to work W = F · dr) of a constraint forcevanishes. More generally, there is no net work from constraints, so i Zi · δri = 0 (whichholds for the actions of surfaces, rolling constraints, and similar

∫things). The Newtonian

equation of motion is pi = Fi + Zi, where Fi encapsulates other forces.

∑Vanishing virtual

work gives ∑(pi

i

− Fi) · δri = 0 (1.64)

which is the D’Alembert principle. This could be taken as the starting principal for classicalmechanics instead of the Hamilton principle of stationary action.

Of course Eq.(1.64) is not fully satisfactory since we are now used to the idea of workingwith generalized coordinates rather than the cartesian vector coordinates used there. So letstransform to generalized coordinates through r ∂

i = ri(q, t), so δri = ri δqj, where again we∂qj

sum over repeated indicies (like j here). This means

∂rFi · δri = Fi · i

δqj∂qj

≡ Qjδqj (1.65)

where we have defined generalized forces

∂rQj ≡ Fi · i

. (1.66)∂qj

We can also transform the pi · δri term using our earlier point transformation results as well

as the fact that d ∂dt

(ri

2

=∂qj

)∂ ri

2

+∂qj∂t

∑∂ ri

k qk = ∂vi∂qj∂qk

. Writing out the index sums explicitly,∂qj

this gives ∑·

∑ ∂r· ipi δri = miri

i i,j

δqj∂qj

=∑i,j

(d ∂

mdt

(r

iri · i dm

∂qj

)− iri ·

(∂ri

dt

))δqj

∂qj

=∑i,j

(d(

∂vmivi

dt· i ∂

m∂qj

)v− ivi · i

δ∂qj

)qj

=∑j

(d ∂

dt

(T ∂

∂qj

)T−)δqj (1.67)

∂qj

for T = 1∑

m2 i iv

2i . Together with the D’Alembert principle, we obtain the final result∑

j

(d ∂

dt

(T ∂

∂qj

)T− Q∂qj− j

)δqj = 0. (1.68)

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We will see momentarily that this result is somewhat more general than the Euler-Lagrangeequations, containing them as a special case.

We will start by considering systems with only holonomic constraints, postponing othertypes of constraints to the next section. Here we can find the independent coordinates qjwith j = 1, . . . , N−k that satisfy the k constraints. This implies that the generalized virtualdisplacements δqj are independent, so that their coefficients in Eq. (1.68) must vanish,

d ∂

dt

(T ∂

∂qj

)T−∂q−Qj = 0 . (1.69)

j

There are several special cases of this result, which we derived from the d’Alembert principle.

1. For a conservative force Fi = −∇iV , then

rQj = − ∇ i

( iV ) · ∂V=

∂qj− (1.70)∂qj

where we assume that the potential can be expressed in the generalized coordinatesas V = V (q, t). Then using L = T − V , we see that Eq. (1.69) simply reproduces the

Euler-Lagrange equations d ∂dt

(L ∂

∂qj

)− L = 0.

∂qj

2. If Qj = − ∂V + d∂qj

∂dt

(V)

for V = V (q, q, t), which is the case for velocity dependent∂qj

forces derivable from a potential (like the electromagnetic Lorentz force), then the

Euler-Lagrange equations d ∂dt

(L ∂

∂qj

)− L = 0 are again reproduced.

∂qj

3. If Qj has forces obtainable from a potential as in case 2, as well as generalized forcesRj that cannot, then

d ∂

dt

(L ∂

∂qj

)L− = Rj (1.71)

∂qj

is the generalization of the Euler-Lagrange equations with non-conservative generalizedforces.

An important example of a nonconservative forces Rj is given by friction.

• Static friction is Fs ≤ Fmaxs = µsFN for a normal force FN.

• Sliding friction is F = −µF vN , so this is a constant force that is always opposite thev

direction of motion (but vanishes when there is no motion).

• Rolling friction is F = −µRFv

N .v

• Fluid friction at a low velocity is F = −bv v =v−bv.

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A general form for a friction force is Fi = −hi(vi)vi (where as a reminder there is novi

implicit sum on i here since we specified i on the right-hand-side). For this form

Rj = −∑ vi

hii

∂r

vi· j v

=∂qj

−∑

ihi

i

∂v

vi· i

. (1.72)∂qj

Simplifying further gives

Rj = −∑ hi

i

2vi

∂v

∂qj

(2i

)= −

∑ vihi

i

∂=

∂qj−∑ vi

i

∂qj

v

∂vi

∫i ∂dvi′hi(vi

′) =0

−v

∂qj

∑i

∫i

dvi′hi(vi

′)0

∂=

F− (1.73)∂qj

where

F =∑∫ vi

dvi′ hi(vi

′) (1.74)0i

is the “dissipation function”. This is a scalar function like L so it is relatively easy to workwith.

Example: Consider a sphere of radius a and mass m falling in a viscous fluid. ThenT = 1m′y2 where m′ < m accounts for the mass of displaced fluid (recall Archimedes princi-

2

ple that the buoyant force on a body is equal to the weight of fluid the body displaces). AlsoV = m′gy, and L = T−V . Here h ∝ y, so F = 3πηay2, where by the constant of proportion-

ality is determined by the constant η, which is the viscosity. From this, d ∂dt

(L ∂∂y

)− L =

∂y−∂F

∂y

gives the equation of motion m′y + m′g = −6πηay. The friction force 6πηay is known asStokes Law. (We will derive this equation for the friction force from first principles lateron, in our discussion of fluids.) This differential equation can be solved by adding a par-ticular solution yp(t) to a solution of the homogeneous equation m′yH + 6πηayH = 0. Forthe time derivatives the results are yp = −m′g/(6πηa) and yH = A exp(−6πηat/m′), wherethe constant A must be determined by an initial condition. The result y = yH + yp can beintegrated in time once more to obtain the full solution y(t) for the motion.

Example: if we add sliding friction to the case of two masses on a plane connected bya spring (considered on problem set #1), then hi = µfmig for some friction coefficient µf ,and

F = µfg(m1v1 +m2v2) = µfg(m1

√x2

1 + y21 +m2

√x2

2 + y22 . (1.75)

If we switch to a suitable set of generalized coordinates qj that simplify

)the equations of

motion without friction, and then compute the generalized friction forces Rj = − ∂F , we can∂qj

get the equations of motion including friction. Further details of how this friction complicatesthe equations of motion were provided in lecture.

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

1.5 Calculus of Variations & Lagrange Multipliers

Calculus of Variations

In the calculus of variations, we wish to find a set of functions yi(s) between s1 and s2 thatextremize the following functional (a function of functions),

J [yi] =

∫ s2

ds f(y1(s), . . . , yn(s), y1(s), . . . , yn(s), s) , (1.76)s1

where for this general discussion only we let yi ≡ dyi rather than dds

. To consider the action ofdt

the functional under a variation we consider yi′(s) = yi(s) + ηi(s) where ηi(s1) = ηi(s2) = 0,

meaning that while the two endpoints are fixed during the variation δyi = ηi, the path inbetween is varied. Expanding the variation of the functional integral δJ = J [yi

′]− J [yi] = 0to 1st order in δyi we have

s2 ∂f0 = δJ =

∫ds

s1

∑i

[δyi

∂f+ δyi

∂yi ∂

]. (1.77)

yi

Using integration by parts on the second term, and the∫vanishing∑ [ of the variation at thes

endpoints to remove the surface term, δJ vanishes when 2 ∂fs1 i

d∂yi− ∂

ds

(f δyi(s) ds =

∂yi

0. For independent variations δyi (for example, after imposing holonomic constrain

)]ts), this

can only occur if∂f d

∂yi−

(∂f

ds= 0 . (1.78)

∂yi

The scope of this calculus of variation result for

)extremizing the integral over f is more

general than its application to classical mechanics.

Example: Hamilton’s principle states that motion qi(t) extremizes the action, so in thiscase s = t, yi = qi, f = L, and J = S. Demanding δS = 0 then yields the Euler-Lagrangeequations of motion from Eq. (1.78).

Example: As an example outside of classical mechanics, consider showing that the shortestdistance between points on a∫ sphere of radius a are great circles. This can be seen by

sminimizing the distance J = 2 ds where for a spherical surface,

s1

ds =√

(dx)2 + (dy)2 + (dz)2 =√a2(dθ)2 + a2 sin2(θ)(dφ)2 (1.79)

since dr = 0. Taking s = θ and y = φ, then

ds = a

√dφ

1 + sin2(θ)

( )2

dθ, (1.80)dθ

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so f =√

˙1 + sin2(θ)φ2. The solution for the minimal path is given by solving d ∂dθ

(f ∂∂ϕ

)− f =

∂ϕ

0. After some algebra these are indeed found to be great circles, described by sin(φ− α) =β cot(θ) where α, β are constants.

Example: Hamilton’s principle can also be used to yield the Hamilton equations of motion,by considering the variation of a path in phase space. In this case∫ t2

δJ [q, p] = δ dt[piqi −H(q, p, t)

]= 0 (1.81)

t1

must be solved with fixed endpoints: δqi(t1) = δqi(t2) = 0 and δpi(t1) = δpi(t2) = 0. Here,the role of yi, of is played by the 2N variables (q1, . . . , qN , p1, . . . , pN). As f = piqi−H, then

d ∂

dt

(f ∂

∂qi

)f− ∂H

= 0 =∂qi

⇒ pi = − , (1.82)∂qi

d ∂

dt

(f ∂

∂pi

)f− ∂H

= 0 =∂pi

⇒ qi = ,∂pi

giving Hamilton’s equations as expected. Note that because f is independent of pi, theterm (∂f/∂pi)δpi = 0, and it would seem that we do not really need the condition thatδpi(t1) = δpi(t2) = 0 to remove the surface term. However, these conditions on the variationsδpi are actually required in order to put qi and pi on the same footing (which we will exploitlater in detail when discussing canonical transformations).

It is interesting and useful to note that D’Alembert’s principle(d ∂

dt

(L ∂

∂qj

)L− 0

∂−Rj =

qj

)δqj (1.83)

is a “differential” version of the equations that encode the classical dynamics, while Hamil-ton’s principle

Lδ =

∫ t2

J dt1

(∂

t

d

∂qj− ∂

dt

(L

0∂

))δqj = (1.84)

qj

(for Rj = 0 where all forces come from a potential) is an integrated version.

Method of Lagrange Multipliers

Next we will consider the method of Lagrange multipliers. For simplicity we will assumethere are no generalized forces outside the potential, Rj = 0, until further notice. Themethod of Lagrange multipliers will be useful for two situations that we will encounter:

1. When we actually want to study the forces of constraint that are holonomic.

2. When we have semi-holonomic constraints.

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Let us consider k constraints for n coordinates, with α ∈ {1, . . . , k} being the indexrunning over the constraints. These holonomic or semi-holonomic constraints take the form

gα(q, q, t) = ajα(q, t)qj + atα(q, t) = 0 (1.85)

where again repeated indices are summed. Thus, gαdt = ajαdqj + atαdt = 0. For a virtualdisplacement δqj we have dt = 0, so ∑n

ajαδqj = 0 , (1.86)j=1

which gives us k equations constraining the virtual displacements. For each equation wecan multiply by a function λα(t) known as Lagrange multipliers, and sum over α, and thecombination will still be zero. Adding this zero to D’Alembert’s principle yields[

d ∂

dt

(L ∂

∂qj

)L− =

∂− λαajα j

qj

]δq 0 (1.87)

where the sums implicitly run over both α and j. Its clear that the Lagrange multiplierterm is zero if we sum over j first, but now we want to consider summing first over α fora fixed j. Our goal is make the term in square brackets zero. Only n − k of the virtualdisplacements δqj are independent, so for these values of j the square brackets must vanish.For the remaining k values of j we can simply choose the k Lagrange multipliers λα to forcethe k square bracketed equations to be satisfied. This is known as the method of Lagrangemultipliers. Thus all square bracketed terms are zero, and we have the generalization of theEuler-Lagrange equations which includes terms for the constraints:

d ∂

dt

(L ∂

∂qj

)L− = λαajα . (1.88)

∂qj

This is n equations, for the n possible values of j, and on the right-hand-side we sum over αfor each one of these equations. The sum λαajα can be interpreted as a generalized constraintforce Qj. The Lagrange multipliers λα and generalized coordinates qj together form n + kparameters, and equation (1.88) in conjunction with gα = 0 for each α from (1.85) togetherform n+ k equations to be solved.

There are two important cases to be considered.

˙1. In the holonomic case, f (q, t) = 0. Here, g = f = ∂fαα α α qj + ∂fα

∂qj, so ajα = ∂fα

∂t. This

∂qjgives

d ∂

dt

(L ∂

∂qj

)L−

∑k ∂fα= λα

∂qj α=1

(1.89)∂qj

for holonomic constraints. The same result can be derived from a generalized Hamil-ton’s principle

t2

J [qj, λα] =

∫(L+ λαfα) dt (1.90)

t1

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

by demanding that δJ = 0. It is convenient to think of −λαfα as an extra potentialenergy that we add into L so that a particle does work if it leaves the surface definedby fα = 0. Recall that given this potential, the Forceq = −∇q(−λαfα) = λα∇qfα,where the derivative ∇qfα gives a vector that is normal to the constraint surface ofconstant fα = 0. This agrees with the form of our generalized force above.

2. In the semi-holonomic case, we just have gα = ajα(q, t)qj+atα(q, t) = 0, with a αα = ∂gj .

∂qj

This gives

d ∂

dt

(L ∂

∂qj

)L−

∑k ∂gα= λα

∂qj α=1

(1.91)∂qj

for semi-holonomic constraints. This result cannot be derived from Hamilton’s principlein general, justifying the time we spent discussing d’Alembert’s principle, which wehave used to obtain (1.91). Recall that static friction imposes a no-slip constraint inthe form of our equation gα = 0. For g ∝ q, the form , ∂g , is consistent with the form

∂q

of generalized force we derived from our dissipation function, ∂F from our discussion∂q

of friction.

We end this chapter with several examples of the use of Lagrange multipliers.

Example: Consider a particle of mass m at rest on the top of a sphere of radius a, as shownabove in Fig. 1.7. The particle is given an infinitesimal displacement θ = θ0 so that it slidesdown. At what angle does it leave the sphere?We use the coordinates (r, θ, φ) but set φ = 0 by symmetry as it is not important. Theconstraint r ≥ a is non-holonomic, but while the particle is in contact with the sphere theconstraint f = r − a = 0 is holonomic. To answer this question we will look for the point

where the constraint force vanishes. Here T = m ˙r2 + r2θ2 and V = mgz = mgr cos(θ) so2

that L = T − V , then d

( )∂

dt

(L ∂∂r

)− L = λ∂f

∂rgives

∂r

mr − ˙mrθ2 +mg cos(θ) = λ, (1.92)

while d ∂dt

(L ∂˙∂θ

)− L = λ∂f

∂θ= 0 gives

∂θ

d.

dt

(mr2θ

)−mgr sin(θ) = 0 (1.93)

This in conjunction with r = a gives 3 equations for the 3 variables (r, θ, λ). Putting themtogether gives r = 0 so r = 0. This means

¨ ˙ma2θ = mga sin(θ), −maθ2 +mg cos(θ) = λ.

˙ ˙Multiply the first of these by θ and integrate over time, knowing that θ = 0 when θ = 0,˙gives θ2 = 2g (1

a− cos(θ)). Thus,

λ = mg(3 cos(θ)− 2) (1.94)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

is the radial constraint force. The mass leaves the sphere when λ = 0 which is whencos(θ) = 2 (so θ ≈ 48o).

3

What if we instead imposed the constraint f ′ = r2 − a2 = 0? If we call its Lagrangemultiplier λ′

′we would get λ′ ∂f = 2aλ =

r′ when r a, so 2aλ′ = λ is the constraint force from

before. The meaning of λ′ is different, and it has different units, but we still have the sameconstraint force.

What are the equations of motion for θ > arccos(

2 ?3

)Now we no longer have the

constraint so

2 d˙mr −mrθ +mg cos(θ) = 0 and ˙mrdt

(2θ)−mgr sin(θ) = 0.

The initial conditions are r1 = a, θ1 = arccos(

2 ˙,3

)r1 = 0, and θ2

1 = 2g from before. Simpler3a

coordinates are x = r sin(θ) and z = r cos(θ), giving

mL = x2 + z2 (1.95)

2−mgz,

so x = 0 and z =

( )−g with initial conditions z1 = 2a

√, x1 =

35a , and the initial velocities3

simply left as z1 and x1 for simplicity in writing (though the actual values follow from˙z1 = − ˙a sin θ1θ1 and x1 = a cos θ1θ1). This means

x(t) = x1(t− t1) + x1, (1.96)g

z(t) = − (t− t1)2 + z1(t2

− t1) + z1, (1.97)

where t1 is the time when the mass leaves the sphere. That can be found from

θ2 2g=

4g(1

a− cos(θ)) =

θsin2

a

( ), (1.98)

2

so t1 =√

a arccos

4g

∫ ( 23) dθ

θ0 sin( θwhere θ angula

) 0 is the small initial r displacement from the top2

of the sphere.

Example: Consider a hoop of radius a and mass m rolling down an inclined plane of angleφ without slipping as shown in Fig. 1.11, where we define the x direction as being parallel tothe ramp as shown. What is the friction force of constraint, and how does the accelerationcompare to the case where the hoop is sliding rather than rolling?

˙ ˙The no-slip constraint means aθ = x, so h = aθ − x = a, which can be made holonomicbut which we will treat as semi-holonomic. Then T = TCM + Trotation = 1

2mx2 + 1 ˙ma2θ2 as

2

Ihoop = ma2. Meanwhile, V = mg(l − x) sin(φ) so that V (x = l) = 0. This means

mL = T − V =

ma2

x2 +2

θ2 +mg(x2

− l) sin(φ). (1.99)

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

Figure 1.11: Hoop rolling on inclined plane

The equations of motion from ddt

(∂L ∂∂x

)− L = λ∂h

∂xand d

∂x∂

dt

(L)

∂˙∂θ− L = λ∂h

∂θ ˙ are∂θ

mx−mg sin(φ) = λ and ma2θ = λa, (1.100)

˙ ¨along with x = aθ. Taking a time derivative of the constraint gives x = aθ, so mx = λ, andx = g sin(φ). This is one-half of the acceleration of a sliding mass. Plugging this back in we

2

find that1

λ = mg sin(φ) (1.101)2

is the friction force in the − ¨x direction for the no-sliding constraint, and also θ = g sin(φ).2a

Example: Consider a wedge of mass m2 and angle α resting on ice and moving withoutfriction. Let us also consider a mass m1 sliding without friction on the wedge and try tofind the equations of motion and constraint forces. The constraints are that y2 = 0 so the

Figure 1.12: Wedge sliding on ice

wedge is always sitting on ice, and y1−y2 = tan(α) so the point mass is always sitting on thex1−x2

wedge. (We will ignore the constraint force for no rotation of the wedge, and only ask aboutthese two.) The kinetic energy is simply T = m1 (x2

2 1 + y21)+ m2 (x2

2 2 + y22), while the potential

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CHAPTER 1. A REVIEW OF ANALYTICAL MECHANICS

energy is V = m1gy1 +m2g(y2 + y0), where y0 is the CM of the wedge taken from above itsbottom. Then L = T − V , with the constraints f1 = (y1 − y2) − (x1 − x2) tan(α) = 0 andf2 = y2 = 0. The equations of motion from the Euler-Lagrange equations with holonomicconstraints are

d ∂L

dt

∂L

∂x1

− ∂f1= λ1

∂x1

∂f2+ λ2

∂x1

=∂x1

⇒ m1x1 = −λ1 tan(α), (1.102)

d ∂L

dt

∂L

∂y1

− ∂f1= λ1

∂y1

∂f2+ λ2

∂y1

=∂y1

⇒ m1y1 +m1g = λ1,

d ∂L

dt

∂L

∂x2

− ∂f1= λ1

∂x2

∂f2+ λ2

∂x2

=∂x2

⇒ m2x2 = λ1 tan(α),

d ∂L

dt ∂y2

− ∂L ∂f1= λ1

∂y2

∂f2+ λ2

∂y2

= 2y2

⇒ m y2 +m2g =∂

−λ1 + λ2,

which in conjunction with y1−y2 = (x1−x2) tan(α) and y2 = 0 is six equations. We numberthem (1) to (6). Equation (6) gives y2 = 0 so (4) gives m2g = λ2 − λ1 where λ2 is the forceof the ice on the wedge and λ1 is the vertical force (component) of the wedge on the pointmass. Adding (1) and (3) gives m1x1 + m2x2 = 0 meaning that the CM of m1 and m2 hasno overall force acting on it.

Additionally, as (5) implies y1 = (x1 − x2) tan(α), then using (1), (2), and (3) we findthe constant force

gλ1 =

1 2

+ tan (α)m1 cos2(α)

. (1.103)m2

With this result in hand we can use it in (1), (2), and (3) to solve for the trajectories. Since

tan(α)x2 = λ1, (1.104)

m2

tan(α)x1 = − λ1,

m1

λ1y1 = g

m1

− ,

the accelerations are constant. As a check on our results, if m2 →∞, then x2 = 0 so indeedthe wedge is fixed; and for this case, x1 = −g sin(α) cos(α) and y1 = −g sin2(α) which bothvanish as α → 0 as expected (since in that limit the wedge disappears, flattening onto theicy floor below it).

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